Primerano - Population Genetics

Question 1

What is the study of the distribution of alleles in populations and the factors that maintain or alter allele frequencies?

Answer 1

Population genetics

Question 2

_______________ allows us to determine the frequency of a given genotype from the frequencies of the alleles of locus in question.

Answer 2

Hardy-Weinberg equation

Question 3

Allele frequencies are determined by: ?

Answer 3

Counting the number of alleles of one type and dividing by the total number of all the different alleles.

Question 4

******What is the equation for the frequencies of two alleles within one gene?

Answer 4

********p + q = 1

p - dominant allele

q - recessive allele

Question 5

What is the equation that yields the frequency of the three possible genotypes from a gene with two alleles?

Hardy-Weinberg Equation

Answer 5

(p + q)² = p² + 2pq + q² = 1, where

p² = frequency of BB genotype

2pq = frequency of Bb genotype

q² = frequency of bb genotype

Question 6

Facts:

• Allele frequencies have clinical significance.
• Since these frequencies can vary from population to population, the risk of carrier status varies (sometimes dramatically) and influences the risk of having an affected offspring.

Question 7

_____________ allele frequencies allow us to calculate the probability of carrier status in different populations.

• Practical approach to risk assessment, where we don’t have molecular means of assessing this

Answer 7

Population-specific

Question 8

What law states that:

• genotypes will be distributed in a population based on the allele frequencies (mathematical relationship)
• the genotype frequencies will remain constant from generation to generation.

Answer 8

The Hardy-Weinberg Law

Question 9

The Hardy Weinberg law only holds true as long as the following 5 conditions are met:

1.

2.

3.

4.

5.

Answer 9

1. The population exhibits random mating.
2. The trait is present in a large population.
3. There is a negligible net mutation rate between the alleles of interest.
4. There is a negligible amount of migration into or out of the population.
5. There is a negligible amount of selection, that is, all genotypes are equally viable and equally fertile.

Question 10

Which condition of the Hardy Weinburg Law?

• Example: if one genotype (bb) rarely mates, then the b allele frequency will decline over time.
• There are three classes of nonrandom matings:
  
  • Consanguineous matings
  • Matings in a stratified population
  • Assortative mating

Answer 10

Random Mating

Question 11

What types of mating results in an increase in homozygous genotypes and a decrease in heterozygous genotypes?

• More aa x aa matings

Answer 11

Consanguineous Matings

• In a population where nonrandom matings predominate, i.e. mostly BB X BB and bb X bb, heterozygotes will not be generated.
• So the homozygotes approach 50%
• AA ~= 0.50, Aa ~= 0, aa ~= 0.50

Question 12

What types of nonrandom matings?

• ____________ have subgroups that do not intermarry. Basis can be culture, economic class, race and religion.
• ___________ is the choice of mate based on phenotype (e.g. tall x tall)

–> In both, heterozygous types are also underrepresented.

Answer 12

• Stratified populations
• Assortative mating

Question 13

What condition of Hardy Weinberg law?

• In a small population: one genotype may be transmitted exclusively to a subsequent generation simply by chance.
  
  • The net effect would be a reduction in the frequency of alleles not present in this genotype.
• This exclusive transmission can happen in any population, but is much less likely to happen in large populations.

Bonus: what can happen in small populations, where alleles often change in frequency over generations?

Answer 13

Trait is present in a Large Population

Bonus: Genetic Drift

Question 14

What term?

• applies to the establishment of new alleles or new allele frequencies in a population and
• Can also apply to the formation of subpopulations isolated from larger population.

Answer 14

Genetic Drift

Question 15

What condition of Hardy Weinberg law?

• Two alleles may differ by a single base
• Their frequencies can change under the following conditions.
  
  • A high mutation rate in favor of forming the b allele will tend to raise its frequency.
  • Alleles are lost when individuals have no offspring. This can be significant if the trait reduces the fitness of the carrier.

Answer 15

Negligible Net Mutation Rate between the B and b alleles

2 Scenarios:

• Forward = reverse rate of mutation, frequencies stay the same
• High forward rate of mutation, leads to low fitness of bb, frequencies stay the same

Question 16

What is the probability of transmitting one’s genes to the next generation?

• For example, fitness (f) for achondroplasia is 20%. This means that 80% of the defective alleles are lost every generation.

Answer 16

Genetic Fitness

Question 17

If the population is in HW Equilibirum, alleles lost by disadvantage must be replaced by spontatneous mutations.

What is the equation for spontaneous mutation rate (m)?

Answer 17

Can be determined from the mutant allele frequency (q) and the selective disadvantage (s) which is related to f:

s = 1 - f

m = s X q

Ex: for achondroplasia,
defective allele frequency = 1/40,000
f = 20%, so s = 80%

m = 0.8 x 1/40,000 = 2 x 10^-5

Question 18

What condition of Hardy Weinberg law?

• Founder effect occurs when a group of colonists do not have the same allele frequencies as their original population or the population they move into.

Answer 18

Negligible amount of migration into/out of the population

Question 19

Example of what phenomenon?

• Venezuelans have higher frequency of Huntington disease than other parts of South America because one or a few individuals (possibly European colonists) had many descendants there.
• So the allele frequency in Venezuela is higher than in Europe and other S.A. countries.

Answer 19

Founder Effect

Question 20

What condition of Hardy Weinberg law?

• This means that all genotypes are equally viable and equally fertile.
• If there is complete selection against aa individuals, only AA and Aa will contribute to the gene pool of the next generation.
• This will reduce q (freq (a)) over time, but will not completely eliminate it.

Answer 20

Negligible amount of selection

Question 21

Determine the frequency of M and N alleles from the information on the genotypes. Use equations (1) and (2).

Given genotype frequencies:MM= 36;MN= 48;NN = 16.
Total sample = 100

p = freq (B) = {2 (# BB) + (# Bb)}/ (2 X Total) (eq 1)

q = freq (b) = {2 (# bb) + (# Bb)}/ (2 X Total) (eq 2)

p = (2 MM + MN)/2 = (72 + 48)/ 200 = 0.6

q = (2 NN + MN)/2 = (32 + 48)/ 200 = 0.4

We can check the frequencies using HW equation

(MN) = 2pq = 2(0.6)(0.4) = .48

Question 22

For X-linked gene in female populations:

Use the standard form of HW (p² + 2pq + q² = 1 )

For male populations, there are only two possible genotypes:

(1) H/(Y) whose genotype frequency = p, and
(2) h/(Y) whose genotype frequency is q

For males, H-W equation simplifies to p + q = 1.

Question 23

_______________ to an individual is an essential part of genetic counseling.

For single gene disorders, assessing probability of recurrence of a given disorder can be fairly straightforward.

Carrier status for some disorders is readily determined by:

(1) direct detection of the wild type and mutant alleles (by PCR, for example) or
(2) detection of closely linked genetic markers, such as RFLPs or HLA markers.

However, for the majority of diseases the carrier status cannot be directly determined.

Answer 23

Determination of risk

Question 24

Risk calculation for Mendelian Traits:

A) if carrier status is known, Punnett square is sufficient.

B) if carrier status is not known, use the population-based frequency+ Punnett square calculation.

Question 25

KNOW!
General Guide to Population Genetics and Recurrence Risk Problems

• Build a pedigree based on family data
• Determine the genotype and probability of genotype for the prospective couple
• Determine the probability of having an affected offspring given genotype (probable) of the parents.
• Combine the three probabilities (for autosomal).
  P(carrier or affected) x P (carrier or affected) x Punnett probability = P (having an affected)
• Always round p to 1 for this course. ***

Question 26

****FOR ALL PROBLEMS ROUND P = 1

Question 27

All Equations provided on the Exam.